Q5 of 23 Page 1

Attempt of the following sub questions:
Solve:

…(i)


Let x - = m


Square both sides


⇒ x2 - 2 + = m2


⇒ x2 + = m2 + 2


Hence equation (i) becomes


9 × (m2 + 2) – (3 × m) – 20 = 0


⇒ 9m2 + 18 – 3m – 20 = 0


⇒ 9m2 – 3m – 2 = 0


⇒ 9m2 – 6m + 3m – 2 = 0


⇒ 3m(3m - 2) + 1(3m – 2) = 0


⇒ (3m + 1)(3m - 2) = 0


⇒ 3m + 1 = 0 and 3m – 2 = 0


⇒ 3m = -1 and 3m = 2


∴ m = and m =


But x - = m


⇒ x - = and x - =


Consider x - =


⇒ =


Cross multiplying


⇒ 3(x2 – 1) = -x


⇒ 3x2 – 3 = -x


⇒ 3x2 + x – 3 = 0


Using formula x =


⇒ x =


⇒ x =


⇒ x =


⇒ x = and x =


Now consider x - =


⇒ =


Cross multiplying


⇒ 3(x2 – 1) = 2x


⇒ 3x2 – 3 = 2x


⇒ 3x2 + 2x – 3 = 0


Using formula x =


⇒ x =


⇒ x =


⇒ x =


⇒ x =


⇒ x =


⇒ x =


⇒ x =


⇒ x = and x =


Therefore x = , x = , x = and x = is the solution


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